Chapter 5

Radon-Nikodym Theorem

Signed measures come up in at least two occasions. First, for any non-negativeµ-measurable function f , the map E 7→

∫Efdµ is a measure. When f changes

sign, this map is still countably additive and it should be a “signed measure”.Second, in Theorem 2.8 it is shown that every positive linear functional on Cc(X)(X is a locally compact Hausdorff space) comes from a regular Borel measure.From the viewpoint of functional analysis, it is desirable to characterize the dualspace of Cc(X) as the space of “signed measures”. We introduce signed measuresin Section 1. Just like the absolute value of a function is non-negative, it is shownthat the “ absolute value ” of a signed measure, its total variation, is a measure.In Section 2 we establish the important theorem of Radon-Nikodym and discusshow to decompose a measure or a signed measure into its absolute continuousand singular parts with respect to another measure. Then we use this theorem toestablish the full Riesz representation theorem in Section 3. Weak∗ convergenceof sequences of signed measures is discussed in Section 4. As an application ofthe Riesz representation theorem we give a characterization of weakly convergentL1-sequences, part of the Dunford-Pettis theorem. Finally, as another applicationof the Riesz representation theorem, we prove Herglotz-Riesz theorem concerningthe boundary trace of a non-negative harmonic function in Section 5.

5.1 Signed Measures

ConsiderM a σ-algebra on the non-empty set X. A map µ :M→ R is called asigned measure if it satisfies

µ(E) =∞∑j=1

µ(Ej), ∀ partitions Ej of E.

Here Ej is called a (measurable) partition of E if they are mutually disjoint,measurable and E = ∪jEj. One immediately deduces that µ(φ) = 0 and µ(X) isfinite for a signed measure. The latter makes even a non-negative signed measure

1

different from a measure, namely it must be a finite measure.Given a signed measure on (X,M), its total variation (measure) is given by

|µ|(E) = sup∑j

|µ(Ej)|: ∀ partitions Ej of E.

Clearly |µ| (E1) ≤ |µ| (E2) if E1 ⊂ E2, both in M.

Proposition 5.1. The total variation of every signed measure on (X,M) is afinite measure on (X,M).

Proof. We need to show

|µ| (E) =∞∑j=1

|µ| (Ej),

whenever Ej is a partition of E.First we establish subaddivity. Let Ak be a partition of E. We have

∑k

∣∣∣µ(Ak)∣∣∣ =

∑k

∣∣∣∣∣µ(Ak ∩

⋃j

Ej

)∣∣∣∣∣≤∑k

∑j

|µ(Ak ∩ Ej)|

=∑j

∑k

|µ(Ak ∩ Ej)|(

use∑k

∑j

αkj =∑j

∑k

αkj, ∀αkj ≥ 0)

≤∑j

|µ| (Ej).(Ak ∩ Ejk is a partition of Ej

)Taking supremum over all Ak,

|µ| (E) ≤∑j

|µ| (Ej).

Next, we show the reverse inequality. If |µ| (Ej) = ∞ for some j, then|µ| (E) ≥ |µ| (Ej) = ∞, the inequality holds. Let |µ| (Ej) < ∞ ∀j. For ev-ery ε > 0, we can find a partition

Ejk

of Ej for each j such that

|µ| (Ej) ≤∑k

∣∣µ (Ejk

)∣∣+ε

2j.

Now, Ejkk,j forms a partition of E. We have∑

j

|µ| (Ej) ≤∑j

∑k

∣∣µ (Ejk

)∣∣+ ε

≤ |µ| (E) + ε

2

and we get the desired inequality after letting ε→ 0.Finally, we show that |µ|(X) is finite. If on the contrary |µ| (X) =∞. By the

lemma below, we can break X into A1 and B1 such that |µ(A1)| , |µ(B1)| ≥ 1. As|µ| (A1 + |µ| (B1) = |µ| (X), at least one of µ(A1) and µ(B1) has infinite measure.Assume |µ| (A1) = ∞. Apply the same argument to A1 to get A2, B2 such that|µ(A2)| , |µ(B2)| ≥ 1. Assume again that A2 has infinite measure. Keep doing

this we get Bj+1 ⊂ Aj, so that Bj ∩Bk = φ, j 6= k. Letting B =⋃j

Bj, we have

µ(B) =∞∑j=1

µ(Bj).

As µ(B) is finite, this infinite series converges, and, in particular, µ(Bj)→ 0, butthis is in conflict with our construction that |µ(Bj)| ≥ 1 for all j.

Lemma 5.2. If |µ| (E) =∞ for some E ∈M, then there are disjoint measurablesets A,B in E satisfying A ∪B = E and |µ(A)| , |µ(B)| ≥ 1.

Proof. For, let t > 0, there is some partition Ej of E such that∞∑j=1

|µ(Ej)| >

t. We fix a large N such thatN∑j=1

|µ(Ej)| > t. Rearrange Ej’s in the order

such that E1, E2, . . . , Em, all µ(Ej) < 0 and Em+1, . . . , EN , µ(Ej) ≥ 0. Then|µ(E1) + · · ·µ(Em)|+ |µ(Em+1) + · · ·+ µ(EN)| > t, and so, either

|µ(E1) + · · ·+ µ(Em)| > t

2

or

|µ(Em+1) + · · ·+ µ(EN)| > t

2.

Assume, say, it is the former. We take A =m⋃j=1

Ej. Then |µ(A)| > t

2. Let

B = E \ A. Then

|µ(B)| = |µ(E)− µ(A)|≥ |µ(A)| − |µ(E)|

≥ t

2− |µ(E)|

≥ 1

if we choose t > 2 and t > 2 |µ(E)|+ 2.

3

The following proposition yields a lot of signed measures from a single mea-sure.

Proposition 5.3. Let µ be a measure on (X,M) and f ∈ L1(µ). Then

(a)

λ(E) =

∫E

fdµ, ∀E ∈M

is a signed measure.

(b) Its total variation is given by

|λ|(E) =

∫E

|f |dµ.

Proof. (a) is immediate. To show (b) let Ej be a measurable partition of E.Then

∑j

|λ(Ej)| =∑j

∣∣∣∣∣∫Ej

fdµ

∣∣∣∣∣≤

∑j

∫Ej

|f |dµ

=

∫E

|f |dµ.

Taking supremum over all measurable partitions, we obtain

|λ|(E) ≤∫E

|f |dµ.

On the other hand, let A = x ∈ E : f(x) ≥ 0 and B = x ∈ E : f(x) < 0. Aand B form a measurable partition of E. By the definition of the total variation,we have

|λ|(E) ≥ |λ(A)|+|λ(B)|=∫E

|f |dµ.

Denote byM(X) orM(X,M) the collection of all signed measures on (X,M).It is a good exercise to verify that M(X) forms a vector space and is completeunder the norm

‖µ‖ = |µ| (X).

M(X,M) is a Banach space. Given a signed measure µ, let

µ+ =1

2(|µ|+ µ), µ− =

1

2(|µ| − µ).

4

We can writeµ = µ+ − µ− and |µ| = µ+ + µ−,

where µ+ and µ− are finite measures. The decomposition of a signed measureinto the difference of two finite measures is called the Jordan decomposition ofthe signed measure. The terminology comes from a corresponding decompositionfor functions of bounded variation.

5.2 Radon-Nikodym theorem

Let µ be a measure and λ a measure or signed measure on the same σ-algebraM. The measure λ is called absolutely continuous with respect to µ, λ << µin notation, if every µ-null set is λ-null. λ is concentrated on a set A ∈ M ifλ(E) = λ(E ∩ A), for all E ∈ M. Note that the set A is not unique. When λ isconcentrated on A, it is also concentrated on any set containing A and any setof the form A \N where N is λ-null. We call two measures/signed measures λ1and λ2 singular to each other, λ1 ⊥ λ2 in notation, if λ1 and λ2 are concentratedrespectively on A and B where A and B are disjoint. Clearly, λ1 ⊥ λ2 if λ1 isconcentrated on A and λ2(A) = 0.

The following proposition can be derived easily from the definitions above andis left for you to prove.

Proposition 5.4. Let µ be a measure and λi measures or signed measures, i =1, 2. Then

(a) λ is concentrated on A ⇒ |λ| is concentrated on A.

(b) λ1 ⊥ λ2 ⇒ |λ1| ⊥ |λ2|.

(c) λ1 ⊥ µ, λ2 ⊥ µ ⇒ λ1 + λ2 ⊥ µ.

(d) λ1 << µ, λ2 << µ ⇒ λ1 + λ2 << µ.

(e) λ << µ ⇒ |λ| << µ.

(f) λ1 << µ, λ2 ⊥ µ ⇒ λ1 ⊥ λ2.

(g) λ << µ, λ ⊥ µ ⇒ λ = 0.

Example 5.1. Consider (R,M) where M is the σ-algebra consisting of allLebesgue measurable sets. For f ∈ L1(R), define

ν(E) =

∫E

f dL1.

5

Then ν << L1. On the other hand, consider

λ =N∑j=1

αjδxj

where x1, . . . , xN ⊂ R and αj ∈ R. Then λ is a signed measure on (R,M) (infact, on (R,PR).) As L1 is concentrated on R, so does it on R \ x1, . . . , xN. Asλ is concentrated on x1, . . . , xN, λ and L1 are singular. By Proposition 5.3, allν’s are singular to λ too.

Absolute continuity and mutual singularity describe two extreme relationsbetween two measures. It is striking that they are sufficient in certain sense forthe description of the relation between two measures. This result is contained inthe following two theorems.

Theorem 5.5 (Lebesgue Decomposition). Let µ be a σ-finite measure andλ a signed measure on (X,M). There exist a unique pair of signed measures(λac, λs), λac << µ and λs ⊥ µ, such that λ = λac + λs.

Theorem 5.6 (Radon-Nikodym Theorem). Let µ be a σ-finite measure and λa signed measure on (X,M) such that λ << µ. There exists a unique h ∈ L1(µ)such that

λ(E) =

∫E

h dµ, ∀E ∈M.

The function h is called the Radon-Nikodym derivative of λ with respect to µand will be denoted by dλ/dµ.

We will prove these two theorems in one stroke. The proof is due to vonNeumann.

Proof. Step 1. Assume that both µ and λ are finite measures. Setting ρ = µ+ λand define a functional on L2(ρ) by

Λϕ =

∫ϕdλ, ϕ ∈ L2(ρ).

By Cauchy-Schwarz inequality,

|Λϕ| ≤∫|ϕ| dλ

≤∫|ϕ| dρ

≤√ρ(X) ‖ϕ‖L2(ρ) ,

6

so Λ is a bounded linear functional on L2(ρ). By the self-duality of L2(ρ), thereexists some g ∈ L2(ρ) such that∫

ϕdλ =

∫ϕg dρ, ∀ϕ ∈ L2(ρ). (5.1)

Taking ϕ = χE, E ∈M, ϕ belongs to L2(ρ) and we have

λ(E) =

∫E

g dρ,

and1

ρ(E)

∫E

g dρ =λ(E)

ρ(E)∈ [0, 1].

By a previous exercise, g ∈ [0, 1] ρ-a.e. By redefining g in a null set, we mayassume g(X) ⊂ [0, 1]. Let

λac(E) = λ(E ∩ A),

andλs(E) = λ(E ∩B),

where A = x ∈ X : g(x) ∈ [0, 1) and B = x ∈ X : g(x) = 1 . We have λ =λac + λs.

We rewrite (5.1) as ∫ϕ(1− g) dλ =

∫ϕg dµ. (5.2)

Taking ϕ = χB, we get

0 =

∫B

dµ = µ(B).

We conclude that λs ⊥ µ, since by definition λs is concentrated on B. Takingϕ = χE(1 + g + · · ·+ gn) in (5.2), we have∫

χE(1− gn+1) dλ =

∫χE(1 + g + · · ·+ gn)g dµ.

The left hand side of this relation satisfies∫χE(1− gn+1) dλ

=

∫E

(1− gn+1) dλ

=

∫E∩A

(1− gn+1) dλ

→ λ(E ∩ A) = λac(E), as n→∞,

7

where the monotone convergence theorem has been used in the last step. On theother hand, the right hand side becomes∫

E

(1 + g + · · ·+ gn)g dµ =

∫E∩A

g1− gn+1

1− gdµ

→∫E∩A

g

1− gdµ as n→∞.

Setting h = g/(1 − g), h is non-negative and belongs to ∈ L1(µ). We concludethat

λac(E) =

∫E

h dµ, ∀E ∈M,

so, λac << µ.

Step 2. Consider the case that µ is a finite measure and λ is a signed measure.Applying what has been proved to λ+ and λ−, the Jordan decomposition of λ,we have

λ+ = λ+ac + λ+s , λ− = λ−ac + λ−s ,

λ+ac, λ−ac << µ, λ+s , λ−s ⊥ µ.

By Proposition 5.4, λ = λac + λs, λac ≡ λ+ac − λ−ac, λs = λ+s − λ−s , and λac << µand λs ⊥ λac. Moreover, if

λ±ac =

∫h± dµ,

λac =

∫h dµ, where h ≡ h+ − h− ∈ L1(µ).

Step 3. Let µ be a σ-finite measure and λ a signed measure. Let Xj be ameasurable partition of X with finite measure. Let µj = µbXj

and λj = λbXj.

By the previous step,

λj = λjac + λjs, λjac << µj, λjs ⊥ µj,

λjac(E) =

∫E

hj dµj =

∫E

hjχXjdµ, hj ∈ L1(µj), hjχXj

∈ L1(µ).

Letting λac =∞∑j=1

λjac and λs =∞∑j=1

λjs, we have

λ = λac + λs, λac << µ, λs ⊥ µ,

8

and

λ(E) =

∫E

h dµ, where h =∞∑j=1

hjχXj.

By Proposition 5.3,

∞ > |λ|(X) =

∫X

|h|dµ,

so h ∈ L1(µ).

Step 4. To finish the proof we establish uniqueness. Suppose there is a pair(λ1, λ2) with λ1 << µ and λ2 ⊥ µ such that

λ = λ1 + λ2.

As λ = λac + λs, λ1 + λ2 = λac + λs, it implies that λ1 − λac = λs − λ2. Asλ1 − λac << µ and λs − λ2 ⊥ µ, by Proposition 5.3, λ1 − λac = λs − λ2 = 0, thatis, λ1 = λac and λ2 = λs.

We deduce a significant result from the Radon-Nikodym theorem.Let µ be a signed measure on (X,M). From the relation |µ(E)| ≤ |µ| (E),

∀E ∈ M, we know that µ << |µ|. By Radon-Nikodym theorem, we can findsome h ∈ L1(|µ|) such that

µ(E) =

∫E

h d |µ| .

From1

|µ| (E)

∫E

h d |µ| = µ(E)

|µ| (E)∈ [−1, 1],

we know that |h| ≤ 1 a.e. We claim that in fact |h| = 1 a.e. For, let

Ar = x ∈ X : |h| (x) < r , r ∈ (0, 1).

For any partition Aj of Ar,

∑j

|µ(Aj)| =∑j

∣∣∣∣∣∫Aj

h d |µ|

∣∣∣∣∣≤∑j

∫Aj

|h| d |µ|

≤ r∑j

|µ| (Aj)

= r |µ| (Ar).

9

Taking supremum over all partitions Aj,

|µ| (Ar) ≤ r |µ| (Ar).

As |µ| (Ar) ≤ |µ| (X) < ∞, this forces Ar has |µ|-measure zero. So, |h| = 1|µ|-a.e. After redefining h on a set of measure zero, we may assume |h| = 1everywhere. Thus we have

Proposition 5.7. Let µ be a signed measure on (X,M).

(a) There exists an h ∈ L1(|µ|), |h| ≡ 1, such that dµ = h d |µ|, that is,

µ(E) =

∫E

h d |µ| , ∀E ∈M.

(b) There are disjoint measurable sets A and B such that

µ+(E) = µ(E ∩ A)

µ−(E) = −µ(E ∩B), ∀E ∈M.

(c) If µ = λ1 − λ2 where λi are measures, then λ1 ≥ µ+ and λ2 ≥ µ−.

Proof. (a) Already done.(b) Let A = x ∈ X : h(x) = 1 and B = x ∈ X : h(x) = −1. Then A∪B =X and

µ+(E) =1

2(|µ| (E) + µ(E))

=1

2

∫E

(1 + h) d |µ|

=

∫E∩A

d |µ| (use 1 + h = 2 on A, 1 + h = 0 on B)

=

∫E∩A

dµ

= µ(E ∩ A).

10

Similarly,

µ−(E) =1

2(|µ| (E)− µ(E))

=1

2

∫E

(1− h) d |µ|

=

∫E∩B

d |µ|

= −∫E∩B

dµ

= −µ(E ∩B).

(c) As µ = λ1 − λ2 ≤ λ1,

µ+(E) = µ(E ∩ A) ≤ λ1(E ∩ A) ≤ λ1(E).

As −µ = λ2 − λ1 ≤ λ2,

µ−(E) = −µ(E ∩B) ≤ λ2(E ∩B) ≤ λ2(E).

The existence of disjoint A and B such that µ+ = µbA and µ− = µbB is calledthe Hahn decomposition of µ. Proposition 5.6(c) shows certain minimal propertyof the decomposition.

5.3 The Dual Space of C0(X)

The space Cc(X) may not be complete under the supnorm. For instance, thefunction e−x

2is the uniform limit of a sequence of functions in Cc(R), but it

is positive everywhere. The space of all bounded, continuous functions on atopological space forms a Banach space where Cc(X) is its subspace. We denotethe closure of Cc(X) in in this space by C0(X). It is a Banach space. In fact,when X is a locally compact Hausdorff space, one can show that it consists of allcontinuous functions that vanish at infinity. More precisely, f belongs to C0(X)if and only if, for every ε > 0, there is a compact set K such that |f | is less thanε outside K. Of course, C0(X) is equal to Cc(X) when X is compact. As Cc(X)is dense in C0(X), the dual space of Cc(X) can be identified with the dual spaceof C0(X).

From now on we take X to be a locally compact Hausdorff topological space.A signed measure µ ∈M(X) = M(X,B) is called regular if every E ∈ B is inner

11

and outer regular, i.e.,

|µ| (E) = inf |µ| (G) : E ⊂ G, G open , and

= sup |µ| (K) : K ⊂ E, K compact .

We use Mr(X) to denote all regular Borel signed measures on (X,B).

Proposition 5.8. (a) Mr(X) forms a closed subspace of M(X) under thenorm ‖µ‖ = |µ| (X),

(b) µ = µ+ − µ−, µ ∈Mr(X) if and only if µ± ∈Mr(X),

(c) For µ ∈Mr(X) and f ∈ L1(|µ|), the measure λ given by

λ(E) =

∫E

f d|µ|,

belongs to Mr(X).

I leave the proof of this proposition as an exercise.Each regular Borel signed measure induces a bounded linear functional on

Cc(X). In fact, we used to define

Λϕ =

∫ϕdµ, ∀ϕ ∈ Cc(X),

when µ is a measure. Now, for a signed measure, we could use Proposition 5.7(a)to define

Λϕ =

∫ϕh d |µ| ,

or,

Λϕ =

∫ϕdµ+ −

∫ϕdµ−.

With this definition,

|Λϕ| ≤∫|ϕh| d |µ| ≤ |µ| (X) ‖ϕ‖∞ ,

so‖Λ‖≤ |µ|(X) = ‖µ‖, (5.3)

holds and, in particular, it shows that Λ ∈ Cc(X)′.Consider the map Φ : Mr(X)→ Cc(X)′ by Φ : µ 7→ Λ. We claim that that Φ

is linear.

Indeed, writing

∫ϕdµ ≡

∫ϕh d |µ|, we first prove that

∫ϕd(µ1 + µ2) =

∫ϕdµ1 +

∫ϕdµ2, ∀µ1, µ2 ∈Mr(X)

12

and ∫ϕd(cµ) = c

∫ϕdµ, ∀c ∈ R.

The second one follows directly from the definition. For the first one, we have∫ϕd(µ1 + µ2) =

∫ϕd(µ1 + µ2)

+ −∫ϕd(µ1 + µ2)

− ,

∫ϕdµ1 =

∫ϕdµ+

1 −∫ϕdµ−1 ,

and ∫ϕdµ2 =

∫ϕdµ+

2 −∫ϕdµ−2 .

It suffices to show that∫ϕd(µ1 + µ2)

+ +

∫ϕdµ−1 +

∫ϕdµ− =

∫ϕd(µ1 + µ2)

− +

∫ϕdµ+

1 +

∫ϕdµ+

2 .

(5.4)Observe that for any two finite measures λ and µ on the same σ-algebra,∫

f d(λ+ µ) =

∫f dλ+

∫f dµ, ∀ bounded measurablef,

Asµ1 + µ2 = (µ1 + µ2)

+ − (µ1 + µ2)− ,

µ1 = µ+1 − µ−1 , and µ2 = µ+

2 − µ−2 ,

we have(µ1 + µ2)

+ + µ−1 + µ−2 = (µ1 + µ2)− + µ+

1 + µ+2 .

Therefore,∫ϕd(µ1 + µ2)

+ +

∫ϕdµ−1 +

∫ϕdµ− =

∫ϕd(µ1 + µ2)

− +

∫ϕdµ+

1 +

∫ϕdµ+

2 ,

and (5.4) follows.Finally, let us show that the map from µ to Λ is norm-preserving. In view of

(5.3), it remains to show ‖Λ‖≥ ‖µ‖. Indeed, from

Λϕ =

∫ϕhdµ

we choose a suitable ϕ to approximate h. Recall that |h|≡ 1 and |µ| is a finitemeasure, by Lusin’s theorem (Theorem 2.12), for each ε > 0, there exists someϕ ∈ Cc(X) satisfying (a) |ϕ|≤ |h|= 1 and (b) |µ|(A) < ε where A = x : ϕ(x) 6=

13

h(x). We have

‖Λ‖ ≥ |Λϕ|

=

∣∣∣∣∫ ϕhd|µ|∣∣∣∣

≥ |µ|(X \ A)− ε≥ |µ|(X)− 2ε.

The desired inequality follows by letting ε→ 0.Summarizing, we have shown that the map µ 7→ Λ is a norm-preserving, linear

map from Mr(X) to C0(X)′.

Theorem 5.9. Let X be a locally compact Hausdorff space. The Φ : Mr(X,B)→C0(X)′ is a norm-preserving, bijective linear map. In other words, the dual spaceof C0(X) is equal to Mr(X,B).

This is the full version of Riesz representation theorem. It is different fromTheorem 2.8 for two points. First, it deals only with bounded linear function-als while Theorem 2.8 deals with positive linear functionals which may not bebounded. Second, since we are only concerned with the dual of Cc(X) only, wedo not consider outer measures here.

Lemma 5.10. Setting as above, there exists a positive linear functional Λ0 onCc(X) satisfying

|Λf | ≤ Λ0(|f |) ≤ ‖Λ‖‖f‖∞, ∀f ∈ Cc(X).

Proof. For f ∈ C+c (X), we define

Λ0f = sup |Λϕ| : |ϕ| ≤ f, ϕ ∈ Cc(X) .

Clearly, Λ0f ≥ 0, Λ0f2 ≥ Λ0f1 if f2 ≥ f1 ≥ 0 and Λ0(cf) = cΛ0f . We claim

Λ0(f1 + f2) = Λ0f1 + Λ0f2, f1, f2 ∈ C+c (X).

for, observe that

|Λϕ| ≤ ‖Λ‖ ‖ϕ‖∞ ≤ ‖Λ‖ ‖f‖∞ , if |ϕ| ≤ f.

Taking supremum over all these ϕ,

|Λ0f | ≤ ‖Λ‖ ‖f‖∞ . (5.5)

In particular, Λ0f is finite. Let f1, f2 ∈ C+c (X). For ε > 0, there are ϕ1, ϕ2,

14

|ϕ1| ≤ f1, |ϕ2| ≤ f2, such that

Λ0f1 ≤ |Λϕ1|+ ε, Λ0f2 ≤ |Λϕ2|+ ε.

Pick σ1, σ2 ∈ 1,−1 such that |Λϕ1| = σ1Λϕ1, |Λϕ2| = σ2Λϕ2, then

Λ0f1 + Λ0f2 ≤ Λ(σ1ϕ1 + σ2ϕ2) + 2ε

≤ Λ0(f1 + f2) + 2ε.

So,Λ0f1 + Λ0f2 ≤ Λ0(f1 + f2).

To get the reverse inequality, let ϕ ∈ Cc(X), |ϕ| ≤ f1 + f2 and set

ϕ1 =f1ϕ

f1 + f2, ϕ2 =

f2ϕ

f1 + f2on V,

and and ϕ1 = ϕ2 = 0 on X \V where V = x ∈ X : (f1 + f2)(x) > 0. Note thatϕ1, ϕ2 ∈ Cc(X),|ϕ1| ≤ f1, |ϕ2| ≤ f2, and ϕ = ϕ1 + ϕ2. We have

|Λϕ| = |Λϕ1 + Λϕ2|≤ |Λϕ1|+ |Λϕ2|≤ Λ0f1 + Λ0f2.

Taking supremum over all these ϕ,

Λ0(f1 + f2) ≤ Λ0f1 + Λ0f2.

For f ∈ Cc(X), we set

Λ0f = Λ0f+ − Λ0f

−.

Using the old trick

(f1 + f2)+ − (f1 + f2)

− = f1 + f2 = f+1 − f−1 + f+

2 − f−2 ,

it is routine to check that Λ0 is our desired positive linear functional on Cc(X).

Now, we can prove Theorem 5.9. To show that Φ is onto, we need to findsome µ that satisfies Φ(µ) = Λ for any given Λ. Indeed, applying Theorem 2.8to Λ0, we can find a Radon measure λ such that

Λ0f =

∫f dλ, ∀f ∈ Cc(X).

Here λ satisfies

15

• λ(E) = inf λ(G) : E ⊂ G,G open , ∀E ⊂ X,

• λ(E) = sup λ(K) : K ⊂ E,K compact if λ(E) <∞, ∀E ∈ B.

By Lemma 5.10,∣∣∣∣∫ f dλ

∣∣∣∣ = |Λ0f | ≤ ‖Λ‖ ‖f‖∞ , ∀f ∈ Cc(X).

From the proof of Theorem 2.8,

λ(X) = sup |Λ0f | : 0 ≤ f ≤ 1 on X .

In view of this, we haveλ(X) ≤ ‖Λ‖ ,

i.e., λ belongs to Mr(X).From the definition of Λ0 we also have

|Λf | ≤ Λ0f+ + Λ0f

−

= ‖f‖L1(λ), ∀f ∈ Cc(X).

As Cc(X) is dense in L1(λ), Λ can be extended to a bounded linear functional onL1(λ). Since λ is a finite measure, by L1-L∞ duality there exists some h ∈ L∞(λ)such that

Λf =

∫fh dλ ∀f ∈ L1(λ),

and the operator norm of Λ as a linear functional on L1(λ) is equal to ‖h‖∞.Using |Λf |≤ ‖f‖L1(λ), we have ‖h‖∞≤ 1. On the other hand, we have

‖Λ‖ = sup|Λϕ|: ϕ ∈ Cc(X), |ϕ|≤ 1

≤∫|h|dλ

≤ λ(X)

≤ ‖Λ‖,

which forces |h|= 1 λ-a.e. and λ(X) = ‖Λ‖. Now we set

µ(E) =

∫E

hdλ, ∀E ∈ B.

By Proposition 5.8, µ ∈Mr(X). By Proposition 5.3, |µ|= λ and so ‖µ‖= ‖Λ‖. Weconclude that Φ is a norm-preserving linear bijection from Mr(X,B) to C0(X)′.The proof of Theorem 5.9 is completed.

16

5.4 Weak∗ Convergence of Measures

Let E be a normed space and E ′ its dual space. A sequence Λk ⊂ E ′ is calledweakly∗ convergent to Λ ∈ E ′ if

Λkx→ Λx, ∀x ∈ E,

as k →∞. Consider Lp(µ), 1 < p <∞, we know that Lp(µ) = Lq(µ)′ where q is

conjugate to p. For fk ∈ Lp, fk∗ f if and only if∫

fkg dµ→∫fg dµ, ∀g ∈ Lq(µ).

This turns out to be the same as fn f . So when 1 < p <∞, weak∗ and weakconvergence are the same, depending on whether to regard Lp(µ) as the “base”space or the dual of Lq(µ).

Applying to the case the space C0(X) where X is a locally compact Hausdorffspace, by the representation theorem

µn∗ µ in Mr(X)

if and only if ∫ϕdµn →

∫ϕdµ, ∀ϕ ∈ Cc(X).

The following result is sometimes called Helly selection theorem.

Theorem 5.11. Let E be a separable normed space. Every bounded sequence inE ′ contains a weakly∗ convergent subsequence.

This theorem can be proved as Theorem 5.8 and we omitted its proof. As aspecial case we have

Corollary 5.12. Let C0(X) be separable. Every sequence µk in Mr(X), ‖µk‖ ≤M for some M , contains a subsequence µkj which µkj

∗ µ for some µ ∈Mr(X).

One can show that the space C0(X) is separable when X is a compact metricspace or it is an open set in Rn.

Recall that even ‖fk‖L1 is uniformly bounded, we cannot always pick a weaklyconvergent subsequence fnj

in L1(µ). As a typical example we may take fksatisfying sptfk = [ak, bk] shrinks to x0 and ‖fk‖L1= 1 for all k. Then nosubsequence of fk converges weakly. However, if we regard an L1-function asa measure by setting dµk = fk dL1, µk now belongs to the larger space Mr(R)

and ‖µk‖ = ‖fk‖L1= 1. By Corollary 5.10, it has weak∗ subconvergence µkj∗

µ. In fact, it is clear that the entire sequence converges weakly∗ to the Diracmeasure δx0 . The lesson is, by enlarging the space from L1(R) to Mr(R), we getsubconvergence.

17

As an application of this property, we prove

Theorem 5.13. Consider L1(Ω) where Ω be a bounded, open set in Rn. Letfk be a bounded sequence in L1(Ω) which is uniformly integrable. Then fkcontains a weakly convergent subsequence in L1(Ω).

In fact, the converse is true, namely, if fk∗ f for some f ∈ L1(Ω), then

fk is bounded in L1(Ω) and uniformly integrable. This necessary and sufficientcondition for the weak subconvergence of an L1-sequence is called Dunford-Pettistheorem.

Proof of Theorem 5.13. It suffices to consider the case fk ≥ 0. Let dµn = fn dLn.Then ‖µk‖ = ‖fk‖L1 is uniformly bounded by some constant M . By Corollary5.10, there exist µkj and µ ∈Mr(Ω) such that∫

ϕdµkj →∫ϕdµ, ∀ϕ ∈ Cc(Ω) as j →∞. (5.6)

For simplicity, we assume the entire sequence converges weaklyˆ*. By Lebesguedecomposition (with respect to Ln),

µ = fLn + ν, f ∈ L1(Ω), ν ⊥ Ln.

It suffices to show that ν ≡ 0 so that∫ϕfk dx→

∫ϕf dx, ∀ϕ ∈ Cc(Ω) = C(Ω). (5.7)

Suppose on the contrary that there is someA ∈ B, ν(A) > 0, such that Ln(A) = 0.Letting a0 ≡ ν(A) = µ(A), by the regularity of the Lebsegue measure and µ, forδ > 0, there is an open G containing A such that

µ(G) < δ,

and there is a compact set K ⊂ A such that

Ln(K) ≥ a02.

For ε < a0/2, by uniform integrability, there is a δ1 > 0 such that∫E

fk dx < ε, ∀E ∈ B, Ln(E) < δ1, ∀k ≥ 1.

We take δ = δ1 and pick ϕ ∈ Cc(Ω) such that ϕ ≡ 1 on K, 0 ≤ ϕ ≤ 1, sptϕ ⊂ G.Then ∫

G

ϕfk dx ≤∫G

fk dx < ε.

18

Letting k →∞, by (5.6) ∫G

ϕdµ ≤ ε.

However, as ϕ ≡ 1 on K, we get

µ(K) ≤∫G

ϕdµ ≤ ε,

contradicting our choice of ε. So ν ≡ 0.We still have to show (5.7) holds for all ϕ ∈ L∞(Ω) = L1(Ω)′. First, we claim

that it holds for all ϕ = χG, where G is open. For, let Gj ↑ G,Gj ⊂⊂ Gj+1 andϕj, Gj < ϕj < Gj+1. The claim follows from the uniform integrability of fk.Next, using outer regularity and uniform integrability we know that (5.7) holds forϕ = χE for E ∈ B. Consequently, it also holds for all simple functions s. Now, letϕ ∈ L∞(Ω), say, |ϕ| ≤M . For ε > 0, let −M −1 = a1 < a2 < · · · < aN = M + 1,∆aj < ε. Define

s =N−1∑j=1

ajχAj, Aj = x ∈ Ω : aj ≤ ϕ(x) < aj+1 .

Then ‖s− ϕ‖∞ < ε, so∣∣∣∣∫ ϕfk −∫ϕf

∣∣∣∣ ≤ ∣∣∣∣∫ (ϕ− s)(fk − f)

∣∣∣∣+

∣∣∣∣∫ s(fk − f)

∣∣∣∣≤ ε× 2M +

∣∣∣∣∫ s(fk − f)

∣∣∣∣ .Letting n→∞,

limk→∞

∣∣∣∣∫ ϕfk −∫ϕf

∣∣∣∣ ≤ 2Mε,

and we finally conclude that (5.7) holds for all ϕ in L∞(Ω).

5.5 Herglotz-Riesz Theorem

We present another application of the Riesz representation theorem.Recall that a harmonic function is the real part of an analytic function and it

satisfiesuxx + uyy = 0

in the plane. Let DR = z : |z| < R, z = x + iy, be the disk of radius R. Itis well-known that given a continuous function g on the boundary |z| = R, g(θ),z = Reiθ, θ ∈ [0, 2π], there is a unique harmonic function u in DR which is equalto g on its boundary. In fact, this harmonic function is given by the Poisson

19

formula

u(z) =1

2π

∫|w|=R

R2 − |z|2

|w − z|2g(θ) dθ. (5.8)

The question is: Instead of continuous data g, what are the most general datathat can be imposed on the boundary of DR to get solvability?

It turns out we have

Theorem 5.14. Let u be a non-negative harmonic function defined in D1 =z : |z| < 1. There exists a unique Radon measure µ on S1 such that∫

u(reiθ)ϕ(θ) dx→∫ϕ(θ) dµ(θ), ∀ϕ ∈ C(S1), (5.9)

as r ↑ 1. In fact,

u(z) =1

2π

∫ 2π

0

1− r2

1− 2r cos(α− θ) + r2dµ(θ), z = reiα, r ∈ (0, 1), α ∈ [0, 2π].

(5.10)

Thus every non-negative harmonic function in D1 has a boundary value whichis a Radon measure. Moreover, every Radon measure on the boundary generatesa harmonic function by the formula (5.10).

Proof. Let C(S1) be the space consists of all 2π–periodic, continuous functions.For 0 < r < 1, set

Λrϕ =

∫ 2π

0

u(reiθ)ϕ(θ) dθ.

We have

|Λrϕ| ≤∫ 2π

0

u(reiθ) |ϕ(θ)| dθ

≤ ‖ϕ‖∞∫ 2π

0

u(reiθ) dθ.

Setting z = 0 in (5.8) where now R = r and g = u(reiθ), we obtain the meanvalue property of the harmonic function

u(0) =1

2π

∫ 2π

0

u(reiθ) dθ.

Therefore,|Λrϕ| ≤ 2πu(0) ‖ϕ‖∞ ,

and‖Λr‖ ≤ 2πu(0) ≡M <∞.

20

By Corollary 5.10 or Theorem 2.8, there exist rj ↑ 1 and a regular Borelmeasure µ such that

Λrjϕ =

∫ 2π

0

u(rjeiθ)ϕ(θ) dθ →

∫ 2π

0

ϕ(θ) dµ(θ), rj ↑ 1.

We will later improve the convergence from a sequence rj ↑ 1 to for all r ↑ 1.Now, for a fixed z = reiα, 0 < r < 1, the function

ϕR(θ) =R2 − r2

|Reiθ − reiα|2=

R2 − r2

1− 2r cos(α− θ) + r2

converges uniformly to

ϕ1(θ) =1− r2

1− 2r cos(α− θ) + r2

as R ↑ 1. It follows that

Λrjϕrj →∫ 2π

0

ϕ1(θ)dµ(θ), rj ↑ 1.

On the other hand, by the Poisson formula,

ΛRφR =

∫ 2π

0

R2 − r2

1− 2r cos(α− θ) + r2u(reiθ) dθ

= 2πu(z).

Thus, we have

2πu(z) =

∫ 2π

0

1− r2

1− 2r cos(α− θ) + r2dµ(θ),

that is, (5.10) holds.Now, let g ∈ C(S1). For a fixed r < 1, by Fubini’s theorem and Poisson’s

formula,∫ 2π

0

u(reiα)g(α) dα =1

2π

∫ 2π

0

∫ 2π

0

1− r2

1− 2r cos(α− θ) + r2g(α) dαdµ(θ)

=

∫ 2π

0

h(reiθ) dµ(θ),

where h is the harmonic function determined by the boundary value g. In case

21

(5.10) holds for some measure µ1 instead of µ, the same reasoning shows that∫ 2π

0

u(reiα)g(α) dα =

∫ 2π

0

h(reiθ) dµ1(θ).

It follows that ∫ 2π

0

h(reiθ) dµ1(θ) =

∫ 2π

0

h(reiθ) dµ(θ).

Letting r ↑ 1, ∫ 2π

0

g(θ) dµ1(θ) =

∫ 2π

0

g(θ) dµ(θ), ∀g ∈ C(S1)

so µ1 = µ. This shows the uniqueness of the boundary trace µ. As every rj ↑ 1contains a subsequence rjk such that Λrkj converges weakly∗ to some boundarytrace which must be µ, we conclude that (5.9) holds.

Comments on Chapter 5. Sections 1–4 are taken from [R]. There are differentproofs of Radon-Nikodym theorem, check them from the web. See [R] for a proofof Lp − Lq duality based on the representation theorem when the measure is σ-finite. The discussion in Section 4 is parallel to Section 6 in Chapter 4 and onemay consult Buttazzo, Gaiquinta and Hildebrandt “One-dimensional VariationalProblems” for a complete proof of Dunford-Pettis theorem. Finally, Herglotz-Riesz theorem is taken from Lax “Functional Analysis”.

22